|
| Let's begin with this example: | LHS | RHS | |
| We start with the original equation: | 2 × t + 31 | = | 13 |
|
Step 1: Since the LHS and RHS are actually the same number, let's make a new equation, in which the LHS and RHS are both equal to this number minus 31... |
(2 × t + 31) − 31 | = | (13) − 31 |
|
...we can then simplify the LHS and RHS to get: |
2 × t | = | -18 |
|
oh, and by the way, 2 × t can be abbreviated to 2t. Generally,
with
algebraic symbols, if you write a symbol right next to a number or
another symbol, without specifying +, ×, −,
÷,
etc.
then
multiplication is assumed.
|
2t | = | -18 |
|
Step 2: Great, so we now know what 2 times t is. But we just want t itself, in other words, 1 times t. So, since we know that 2t is equal to -18, therefore 2t divided by 2, must be equal to -18 divided by 2. |
2t / 2 | = | -18 / 2 |
|
of course, 2t / 2 (which equals
t ×
2/2) is just equal to t.
Likewise, any
number divided by itself equals 1.
Hence our answer: |
t | = | -9 |
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